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"The combined results of several people working together is often much more effective than could be that of an
individual scientist working alone." - John Bardeen, only person to win Nobel Prize in physics twice.

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Week 1

Opgave 1a.Let \(v \in V - \{0\}\). Show that there exists a \(\phi \in V^{\ast}\) such that \(\phi(v)\neq 0\).Bewijs. Since \(v \neq 0\), clearly there exists a basis element \(b_0\) such that \(b_0^{\ast}(v) \neq 0\). Since \(b_0^{\ast}\) is in the dual vector space, we are done.

Opgave 1b.Show that there exists a canonical injective linear map \(i:V \to V^{**}\).Bewijs. Define this map as \(i:v \mapsto V^*\to k : \phi \mapsto \phi(v)\). Linearity follows from the linearity from elements of \(V^*\) and from the linearity of elements of \(V^{**}\). Suppose that \(v,w \in V\) such that \(i(v)=i(w)\). In other words, for all \(\phi \in V^*\)\(\phi(v)=\phi(w)\). As \(\phi\) is linear, \(\phi(v-w)=0\). But we know from 1a, that there exist a \(\phi\) such that \(\phi(v-w) \neq 0\) if \(v-w \neq 0\). Therefore \(v-w\) must be zero, and so \(v=w\), as required.

Opgave 1c.Show that \(i\) is an isomorphism if and only if \(V\) is finite dimensional.Bewijs. (\(\Leftarrow\)) Since \(V\) is finite dimensional, \(\dim(V)=\dim(V^{**})\). And as \(i\) is injective, and the dimensions of domain and codomain are the same, \(i\) must be bijective, and therefore it is an isomorphism.

Opgave 2a.Show that \(\beta^* := {b^* : b \in \beta} \subset V^*\) is a linearly independent set.Bewijs. Assume there are scalars such that \[\sum_{b^*\in\beta^*} a_{b^*}b^* =0.\] Well, then since \(b^*\) is in the dual vector space, for all \(c\in \beta^*\), \[a_{c^*}=\sum_{b^*\in\beta^*} a_{b^*}b^*(c) =0(c)=0.\] So all scalars must be zero, as required.

Opgave 2b.Show that \(\beta^*\) is a basis of \(V^*\) if and only if \(V\) is finite dimensional.Bewijs. (\(\Leftarrow\)) If \(V\) is finite dimensional, then for every \(\phi \in V^*\) we’ve got, \[\phi = \sum_{i=1}^{n} \phi(b_i)b_i^*,\] as \[\phi(b_j) = \sum_{i=1}^{n} \phi(b_i) b_i^* (b_j)\] for all \(b_j \in \beta\).